Cell Potential and Free Energy
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AP Chemistry › Cell Potential and Free Energy
A galvanic cell is assembled under standard conditions with the anode labeled as an Al(s)|Al$^{3+}$(aq) electrode and the cathode labeled as a Ag$^+$(aq)|Ag(s) electrode. A student claims the overall cell reaction is spontaneous. Which statement correctly describes the sign of $E^\circ_{\text{cell}}$ and spontaneity for the cell as labeled?
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous because Al is a more active metal than Ag.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
Explanation
This question tests understanding of cell potential and free energy. When the anode and cathode are already labeled, we know that oxidation occurs at the anode (Al → Al³⁺) and reduction occurs at the cathode (Ag⁺ → Ag). Since aluminum has a very negative reduction potential (around -1.66 V) and silver has a positive reduction potential (around +0.80 V), the cell potential E°cell = E°cathode - E°anode will be positive. A positive E°cell indicates that the reaction is spontaneous as written. A common error is assuming that just because aluminum is more active, the reaction direction could be reversed, but the electrode labels define the reaction direction. The key principle is that a positive E°cell always corresponds to a spontaneous galvanic cell reaction.
A student builds a cell under standard conditions using a Sn(s)|Sn$^{2+}$(aq) half-cell and a Ag(s)|Ag$^+$(aq) half-cell. The standard reduction potentials are: Sn$^{2+}$ + 2e$^-$ $\rightarrow$ Sn(s), $E^\circ = -0.14,\text{V}$; Ag$^+$ + e$^-$ $\rightarrow$ Ag(s), $E^\circ = +0.80,\text{V}$. The student writes the overall reaction as Sn$^{2+}$(aq) + 2Ag(s) $\rightarrow$ Sn(s) + 2Ag$^+$(aq). Which statement is correct about this written reaction under standard conditions?
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous because Ag(s) is a good conductor.
Explanation
This question tests understanding of cell potential and free energy. The written reaction shows Sn²⁺ being reduced to Sn and Ag being oxidized to Ag⁺, which is the reverse of the spontaneous direction. With Sn²⁺/Sn at E° = -0.14 V and Ag⁺/Ag at E° = +0.80 V, this reverse reaction would have E°cell = E°cathode - E°anode = (-0.14 V) - (+0.80 V) = -0.94 V. A negative E°cell indicates the reaction is nonspontaneous as written. The error would be assuming that because silver is a good conductor (option E), this affects the thermodynamics, but electrical conductivity doesn't determine reaction spontaneity. The key insight is that reversing a spontaneous galvanic reaction always produces a nonspontaneous reaction with negative E°cell.
A galvanic cell is constructed with a Zn(s)|Zn$^{2+}$(aq) half-cell and a Cu(s)|Cu$^{2+}$(aq) half-cell under standard conditions. A simplified standard reduction potential table is provided:
- Cu$^{2+}$(aq) + 2e$^-$ → Cu(s) $E^\circ = +0.34,\text{V}$
- Zn$^{2+}$(aq) + 2e$^-$ → Zn(s) $E^\circ = -0.76,\text{V}$
Electrons are observed to flow through the external circuit from the Zn electrode to the Cu electrode. Based on this information, which statement is correct about the cell reaction under standard conditions?
The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive.
The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.
The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.
Explanation
This question assesses the skill of cell potential and free energy. In a galvanic cell, the half-cell with the more positive reduction potential serves as the cathode for reduction, while the half-cell with the less positive (or more negative) reduction potential serves as the anode for oxidation. The standard cell potential, E°cell, is determined by subtracting the standard reduction potential of the anode from that of the cathode. A positive E°cell indicates that the cell reaction is spontaneous under standard conditions, as it corresponds to a negative Gibbs free energy change. One tempting distractor is choice B, which incorrectly assumes that a spontaneous reaction must have a negative E°cell, stemming from the misconception that individual negative reduction potentials directly make the cell potential negative. A transferable strategy is to always calculate E°cell as E°cathode minus E°anode and recognize that a positive value confirms a spontaneous galvanic process.
A galvanic cell is assembled from the following two standard reduction half-reactions:
- Al$^{3+}$(aq) + 3e$^-$ → Al(s) $E^\circ = -1.66,\text{V}$
- I$_2$(s) + 2e$^-$ → 2I$^-$(aq) $E^\circ = +0.54,\text{V}$
The Al(s)|Al$^{3+}$ half-cell is connected as the cathode and the I$_2$/I$^-$ half-cell is connected as the anode. Under standard conditions, which statement is correct about the overall cell reaction as assembled?
The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is zero.
The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.
The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive.
Explanation
This question assesses the skill of cell potential and free energy. When half-cells are connected with specified anode and cathode, compare their reduction potentials to find E°cell = E°cathode - E°anode. If E°cell is negative, the reaction is nonspontaneous, indicating a positive ΔG° and the need for external energy. In this assembly, Al's lower E° as cathode versus I2 as anode gives negative E°cell, making it nonspontaneous. Choice A is tempting but wrong, assuming spontaneity with negative E°cell, from the misconception that labeling reverses the natural potential without affecting the sign calculation. A transferable strategy is to always use the formula E°cell = E°red(cathode) - E°red(anode) and note that a negative value means nonspontaneity in the assembled configuration.
A student designs a cell under standard conditions using the half-cells Zn(s)|Zn$^{2+}$(aq) and Ni(s)|Ni$^{2+}$(aq). Standard reduction potentials: Zn$^{2+}$ + 2e$^-$ 6 Zn(s), $E^\circ = -0.76,\text{V}$; Ni$^{2+}$ + 2e$^-$ 6 Ni(s), $E^\circ = -0.25,\text{V}$. If Zn is the cathode and Ni is the anode (as labeled by the student), which statement is correct about the cell as labeled?
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
Explanation
This question assesses the skill of cell potential and free energy. Labeled Zn cathode (-0.76 V), Ni anode (-0.25 V), but Ni has higher E°. E°cell = -0.76 V - (-0.25 V) = -0.51 V, negative. This means nonspontaneous as labeled. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, resulting in positive E°cell error. Use labels; a positive E°cell indicates a spontaneous galvanic process.
A student connects two half-cells under standard conditions: Cu(s)|Cu$^{2+}$(aq) and Ag(s)|Ag$^+$(aq). Standard reduction potentials: Ag$^+$ + e$^-$ 6 Ag(s), $E^\circ = +0.80,\text{V}$; Cu$^{2+}$ + 2e$^-$ 6 Cu(s), $E^\circ = +0.34,\text{V}$. If the student claims electrons flow from Ag to Cu in a spontaneous galvanic cell, which statement is correct about the cell as the student describes it?
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
Explanation
This question assesses the skill of cell potential and free energy. Claimed electron flow from Ag to Cu means Ag anode (+0.80 V), Cu cathode (+0.34 V), but Ag has higher E°. E°cell = +0.34 V - (+0.80 V) = -0.46 V, negative. This indicates nonspontaneity as described. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to positive E°cell wrongly. Base on claimed flow; a positive E°cell indicates a spontaneous galvanic process.
A galvanic cell is built under standard conditions from a Mg(s)|Mg$^{2+}$(aq) electrode and a Cd(s)|Cd$^{2+}$(aq) electrode. The standard reduction potentials are: Mg$^{2+}$ + 2e$^-$ $\rightarrow$ Mg(s), $E^\circ = -2.37,\text{V}$; Cd$^{2+}$ + 2e$^-$ $\rightarrow$ Cd(s), $E^\circ = -0.40,\text{V}$. The student labels Cd as the anode and Mg as the cathode. Based on the provided $E^\circ$ values, which statement is correct about the cell reaction as labeled?
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous because Mg has the more negative reduction potential.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
Explanation
This question tests understanding of cell potential and free energy. The student labeled Cd as anode (oxidation) and Mg as cathode (reduction), meaning the reaction would be Cd → Cd²⁺ + 2e⁻ and Mg²⁺ + 2e⁻ → Mg. This gives E°cell = E°cathode - E°anode = (-2.37 V) - (-0.40 V) = -1.97 V. A negative E°cell indicates the reaction is nonspontaneous as labeled. The error in option E is thinking that having the more negative reduction potential automatically makes Mg the anode, but the student's labeling forces Mg to be the cathode, creating a nonspontaneous cell. The key lesson is that incorrect electrode labeling can result in a nonspontaneous cell with negative E°cell.
A galvanic cell is constructed with a Zn(s)|Zn$^{2+}$(aq) half-cell and a Cu(s)|Cu$^{2+}$(aq) half-cell under standard conditions. A simplified table of standard reduction potentials is provided: Cu$^{2+}$ + 2e$^-$ 6 Cu(s), $E^\circ = +0.34,\text{V}$; Zn$^{2+}$ + 2e$^-$ 6 Zn(s), $E^\circ = -0.76,\text{V}$. If the cell operates as written (Zn electrode is the anode and Cu electrode is the cathode), which statement is correct about spontaneity and the sign of $E^\circ_{\text{cell}}$?
The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
Explanation
This question assesses the skill of cell potential and free energy. The standard cell potential, E°cell, is determined by subtracting the standard reduction potential of the anode from that of the cathode. Here, the Cu half-cell has a higher reduction potential (+0.34 V) compared to Zn (-0.76 V), so when Cu is the cathode and Zn is the anode, E°cell = +0.34 V - (-0.76 V) = +1.10 V, which is positive. A positive E°cell indicates that the redox reaction is spontaneous under standard conditions, as it corresponds to a negative ΔG°. A tempting distractor is choice D, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to a reversed and negative E°cell. Always identify the cathode as the half-cell with the higher reduction potential for spontaneous reactions; a positive E°cell indicates a spontaneous galvanic process.
A student builds a cell under standard conditions using the half-cells Pb(s)|Pb$^{2+}$(aq) and Cu(s)|Cu$^{2+}$(aq). Standard reduction potentials: Cu$^{2+}$ + 2e$^-$ 6 Cu(s), $E^\circ = +0.34,\text{V}$; Pb$^{2+}$ + 2e$^-$ 6 Pb(s), $E^\circ = -0.13,\text{V}$. If the student incorrectly labels Pb as the cathode and Cu as the anode, which statement is correct about the cell as labeled?
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
Explanation
This question assesses the skill of cell potential and free energy. Labeled Pb cathode (-0.13 V), Cu anode (+0.34 V), but Cu has higher E°. E°cell = -0.13 V - (+0.34 V) = -0.47 V, negative. This means nonspontaneous as labeled. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to positive E°cell incorrectly. Use labels to compute; a positive E°cell indicates a spontaneous galvanic process.
A student connects a Fe(s)|Fe$^{2+}$(aq) half-cell to a Ag(s)|Ag$^+$(aq) half-cell under standard conditions. The standard reduction potentials are: Fe$^{2+}$ + 2e$^-$ → Fe(s), $E^\circ=-0.44\ \text{V}$; Ag$^+$ + e$^-$ → Ag(s), $E^\circ=+0.80\ \text{V}$. If electrons flow through the external circuit from the Fe electrode to the Ag electrode, which statement is correct about the overall cell reaction?
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ equals zero.
The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
Explanation
This question tests understanding of cell potential and free energy. When electrons flow from Fe to Ag through the external circuit, Fe is being oxidized (anode) and Ag⁺ is being reduced (cathode), which makes sense because Ag⁺/Ag has the more positive reduction potential (+0.80 V) compared to Fe²⁺/Fe (-0.44 V). The standard cell potential is E°cell = E°cathode - E°anode = +0.80 V - (-0.44 V) = +1.24 V. A positive E°cell indicates the reaction is spontaneous under standard conditions. Students might incorrectly think that because Fe has a negative reduction potential, the overall reaction must be nonspontaneous, but this confuses the individual electrode potential with the cell potential. The key strategy is to identify which electrode serves as cathode (more positive E°) and calculate E°cell = E°cathode - E°anode.